import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;

/**
 * Created with IntelliJ IDEA.
 * Description:
 * User: 26727
 * Date: 2024-01-28
 * Time: 15:55
 */
public class MyBinaryTree {
    static class TreeNode {
        public char val;//数据域
        public TreeNode left;//左孩子地址
        public TreeNode right;//右孩子地址
        public TreeNode(char val) {
            this.val = val;
        }
    }

    public TreeNode root;

    public TreeNode createTree() {
        TreeNode A = new TreeNode('A');
        TreeNode B = new TreeNode('B');
        TreeNode C = new TreeNode('C');
        TreeNode D = new TreeNode('D');
        TreeNode E = new TreeNode('E');
        TreeNode F = new TreeNode('F');
        TreeNode G = new TreeNode('G');
        TreeNode H = new TreeNode('H');
        A.left = B;
        A.right = C;
        B.left = D;
        B.right = E;
        C.left = F;
        C.right = G;
        E.right = H;
        this.root = A;
        return root;
    }
    //前序遍历 根 左子树 右子树
    public void preOrder(TreeNode root) {
        if(root == null) {
            return;
        }

        System.out.print(root.val+" ");
        preOrder(root.left);
        preOrder(root.right);
    }
    //中序 左根右
    public void inOrder(TreeNode root) {
        if(root == null) {
            return;
        }

        inOrder(root.left);
        System.out.print(root.val+" ");
        inOrder(root.right);
    }
    //后序 左右根
    public void postOrder(TreeNode root) {
        if(root == null) {
            return;
        }

        postOrder(root.left);
        postOrder(root.right);
        System.out.print(root.val+" ");
    }

    //获取树中节点的个数
    //子问题思想
    //时间复杂度O(n)和空间复杂度O(logN)[树的高度]
    public int size(TreeNode root) {
        int count = 0;
        if(root == null) {
            return 0;
        }
        int leftSize = size(root.left);
        int rightSize = size(root.right);
        count = leftSize + rightSize + 1;
        return count;
    }

    //遍历思想
       public int NodeSize = 0;//成员变量（可参考写两个方法，算一次没问题）
    //如果上面加static，如果有多个树，就会有问题，不过NodeSize调用就会又错了
       public int size1(TreeNode root) { //也可以不要返回值
       if(root == null) {
           return 0;
       }
       NodeSize++;
       size1(root.left);
       size1(root.right);
       return NodeSize;
    }

    //获取叶子结点的个数
    public int getLeafNodeCount(TreeNode root) {
        int count = 0;
        if(root == null) {
            return 0;
        }
        if(root.left == null && root.right == null) {
            return 1;
        }
        int leftSize = getLeafNodeCount(root.left);
        int rightSize = getLeafNodeCount(root.right);
        count = leftSize + rightSize;
        return count;
    }

    public int LeafSize;
    public int getLeafNodeCount1(TreeNode root) {
        if(root == null) {
            return 0;
        }
        if(root.left == null && root.right == null) {
            LeafSize++;
        }
        getLeafNodeCount1(root.left);
        getLeafNodeCount1(root.right);
        return LeafSize;
    }

    //获取第k层节点个数
    public int getKLevelNodeCount(TreeNode root,int k) {
        if(root == null) {
            return 0;
        }
        if(k == 1) {
            return 1;
        }
        int leftSize  = getKLevelNodeCount(root.left,k-1);
        int rightSize = getKLevelNodeCount(root.right,k-1);
        return leftSize+rightSize;
    }

    //获取二叉树的高度 时间复杂度O(n) 空间复杂度O(树的高度)
    public int getHeight(TreeNode root) {
        if(root == null) {
            return 0;
        }
        int leftHeight = getHeight(root.left);
        int rightHeight = getHeight(root.right);
        return (leftHeight >= rightHeight) ? (leftHeight+1) : (rightHeight+1);
    }

    // 检测值为value的元素是否存在
    public TreeNode find(TreeNode root, int val) {
        if(root == null) {
            return null;
        }
        if(root.val == val) {
            return root;
        }

        /*TreeNode leftTree = find(root.left,val);
        TreeNode rightTree = find(root.right,val);
        return (leftTree!=null) ? leftTree : rightTree;*/

        TreeNode leftTree = find(root.left,val);
        if(leftTree != null) {
            return leftTree;
        }
        TreeNode rightTree = find(root.right,val);
        if(rightTree != null) {
            return rightTree;
        }
        return null;
    }

    //层序遍历
    public void levelOrder(TreeNode root) {
        if(root == null) {
            return;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while(!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            System.out.print(cur.val+" ");
            if(cur.left != null) {
                queue.offer(cur.left);
            }
            if(cur.right != null) {
                queue.offer(cur.right);
            }
        }
    }

    /*
    public List<List<Integer>> levelOrder1(TreeNode root) {
        List<List<Integer>> list = new ArrayList<>();
        if(root == null) {
            return list;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while(!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> list1 = new ArrayList<>();
            while(size != 0) {
                TreeNode cur = queue.poll();
                //System.out.print(cur.val + " ");
                list1.add(cur.val);
                size--;
                if (cur.left != null) {
                    queue.offer(cur.left);
                }
                if (cur.right != null) {
                    queue.offer(cur.right);
                }
            }
            list.add(list1);
        }
        return list;
    }*/

    //判断一棵树是否为完全二叉树[运用队列很巧妙]
    public boolean isCompleteTree(TreeNode root) {
        if(root == null) {
            return true;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while(!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            if(cur != null) {
                queue.offer(cur.left);
                queue.offer(cur.right);
            }else {
                break;
            }
        }
        while(!queue.isEmpty()) {
            TreeNode tmp = queue.poll();
            if(tmp != null) {
                return false;
            }
        }
        return true;
    }

}
